Idioms and Compositionality

Dag Westerståhl March 31, 1999

Abstract

A standard view is that idioms present problems for compositionality. The question of compositionality, however, should be posed for a semantics, not for individual phrases. The paper focuses on the idiom extension problem: Suppose in a given language a certain phrase acquires the status of an idiom. How can the syntax and semantics be extended to accommodate the idiom, while preserving desirable properties such as compositionality? Various ways to achieve such extensions are discussed within an abstract algebraic framework for language due to Wilfrid Hodges.

Contents

1 Introduction 2 2 A methodological point 2 3 Real issues about idioms and compositionality 4 4 The idiom extension problem 4 5 An algebraic framework 5 6 Atomic extensions 7 7 Paraphrase semantics 8 8 How to handle new idiomatic meanings 8 9 Idioms with syntactic structure 10 10 Discussion 11 11 Further directions 13

1 Introduction

If there is one piece of received wisdom in the linguistic community about idioms, it probably is that they spell trouble for the principle of compositionality. After all, the meaning of a typical complex idiom is not determined by the meaning of its parts and the way they are composed, right? If you know English, and in particular the meaning of kick and the bucket, but are unfamiliar with the idiom, there is no way you can compute the (idiomatic) meaning of kick the bucket. Like lexical items, idioms have to be learned one by one. Yet they appear to have syntactic structure, so compositionality is in trouble. Or is it?

In the present paper I want to do two things. First, I claim that the received view as commonly expressed is misleadingly put, and may confuse the issues. Second, a handy algebraic framework (due to Wilfrid Hodges) will be used for a first stab at approaching some of the formal semantic issues that do exist concerning idioms and compositionality. Details and proofs must be deferred to another occasion; here I try to state the main ideas.

There is no way I could do justice to the vast linguistic literature about idioms. Instead I will take one paper, Nunberg, Sag and Wasow [4], as background, because it surveys a fair amount of that literature, and furthermore, it propounds a general view of idioms that seems reasonably widespread, at least in its broad outlines, these days. More precisely, Nunberg et. al amply exemplify with many quotations what I called the received view about idioms and compositionality, and for some idioms such as kick the bucket, they too subscribe to that view. On the other hand, for many other idioms, of which we can take pull strings as a prototype, they claim that compositionality does hold, provided one realizes that the parts or ‘chunks’ of such idioms also have idiomatic meanings.

2 A methodological point

A simple methodological point needs to be made, I think, in connection with the received view about idioms: While it makes good sense to ask if a semantics is compositional or not, it makes no sense to ask the same question about a particular phrase.

Consider how the received view might be put forward, indeed how it might seem self-evident: The meaning of kick the dog, say μ(kick the dog), is a function, say r, of the meanings of kick and the dog:

(1) μ(kick the dog) = r(μ(kick), μ(the dog)).

But, for the idiomatic reading,

(2) μ(kick the bucket) ≠ r(μ(kick), μ(the bucket)).

Yes, but so what? Compositionality only asks that some function determine the meaning, not that this function be r. Simply redefining r for the idiomatic case will do the trick.

This is trivial, but ask any semanticist what the principle of compositionality says, and you will most likely get something like

(C) The meaning of a complex expression is determined by the meanings of its parts and the mode of composition.

And here ‘is determined’ is taken as ‘is a function of’, possibly with some added requirement of computability of that function. But that extra requirement makes no difference to the present issue. One idiomatic exception (hence any finite number) can always be taken care of, it seems.

Perhaps it is retorted that the claim about kick the bucket is in fact intended as a counterexample to the compositionality of a particular semantics. This could be spelled out as follows. Consider the relation ≡_{μ} of synonymy: p ≡_{μ} q iff μ(p) = μ(q). Assume that bucket ≡_{μ} pail. Then

(3) lift the bucket ≡_{μ} lift the pail,

as compositionality would have it, but

(4) kick the bucket ≢_{μ} kick the pail.

contradicting functionality.

This might look better, but actually it only reveals another trivial point. Meanings cannot be assigned to surface manifestations, because then kick the bucket would be ambiguous. Now, ambiguity is not our problem here. Any statement of the compositionality principle already presupposes single-valuedness of meaning. And indeed, as writers about idioms generally acknowledge, on the level where meaning is assigned the idiomatic version of kick the bucket must be distinct from the ordinary version, if only by the presence of an idiomatic ‘marker’.

But once this is acknowledged, the argument loses all force. For then, the ‘modes of composition’ in (1) and (2) are distinct, so there is no reason to expect the same function to operate in both cases. Similarly, on both sides of the synonymy sign in (3), and on the right hand side in (4), we have one ‘mode of composition’, but on the left hand side in (4) we have another, so no violation of compositionality occurs.

Perhaps what the received view really comes down to, in the case at hand, is rather something like this: There is a familiar way to compose a transitive verb meaning with a NP meaning to form a VP meaning, but (a) that way is not used to derive the idiomatic meaning of kick the bucket, and (b) no other familiar way to compose the meanings of kick and the bucket gives the correct result. The familiarity of the function or rule is required to explain how we can figure out, or know, or understand the meaning of a complex phrase. This question is worth discussing, but it is hardly one about compositionality.

Likewise, when Nunberg et al. (appear to) claim that for idioms like pull strings we actually can use the familiar meaning function, given that in the idiom, pull and string have non-standard meanings, this is an interesting issue, but it is still misleading to call those idioms ‘compositional’ and contrast them with ‘non-compositional’ ones. The question is rather: Is there a reasonable compositional semantics for a language containing both kinds of idioms?

3 Real issues about idioms and compositionality

Assume again that bucket ≡_{μ} pail, and suppose we have agreed on a particular ‘mode of composition’, or rule, or marking, that is at work in the idiomatic version of kick the bucket. Can we then avoid, except by ad hoc stipulation, that the same ‘mode of composition’ is applied to yield also an idiomatic version of kick the pail? If not, then, by compositionality, the latter has to mean die as well.

This is a real issue, an instance of the overgeneration problem. Note the difference from the alleged argument above. There we (falsely) claimed to have a counter-instance to compositionality by ignoring that two distinct ‘modes of composition’ were involved. Here we acknowledge the need for an ‘idiomatic mode of composition’, and ask if our analysis, together with compositionality, forces us to countenance idiomatic expressions that do not exist.

The overgeneration problem has been widely discussed in the literature. Other debated issues concern the fact that certain syntactic operations apply to some idioms but not to others. For example,

(5) The bucket was kicked by John yesterday cannot mean that John died yesterday: this idiom does not allow passivization. A similar point concerns anaphoric reference:

(6) Andrew kicked the bucket in June last year, and a month later, Jane kicked it too.

Here too, the idiomatic meaning is hard or impossible to get. But with the literal reading (5) and (6) are both fine (if a bit odd). For the idiom pull strings, on the other hand, both of these operations are all right:

(7) Strings were pulled to secure Henry his position.

(8) Kim’s family pulled some strings on her behalf, but they weren’t enough to get her the job. ([4], p. 502.) One suggestion here is that kick the bucket is somehow an atomic expression,

in contrast with pull strings. This suggestion is worth exploring. But a different idea is that the former idiom stands for a one-place predicate, the latter for a two-place one, so it is natural that passivization and anaphoric reference to the object position apply precisely in the latter case. Then the two idioms could both have syntactic structure but semantics would explain the difference between them.

I think there is some point in discussing such issues also in the abstract, independently of the details of current theories of syntax and semantics. In the rest of this paper I will indicate how to do this within a general algebraic framework for language.

4 The idiom extension problem

Let a language with a compositional semantics be given. Suppose a complex expression in that language acquires, for reasons we need not go into, an idiomatic meaning. How can the given syntax and semantics be extended in a natural way, which accounts for (some of) the behaviour of the new idiom and still preserves compositionality, and perhaps other desirable properties too?

This is the situation I will focus on. It is clear and simple enough and, I think, one we should be able to give a general account of, abstracting from the details of the history of the particular idiom.

5 An algebraic framework

The following definitions are essentially from Hodges [1]. The framework is a much simplified version of the term algebra account of syntax and semantics first introduced by Montague and developed in, for example, Janssen [3].

5.1 Definition.

• A grammar

(E, A, \underline{α})_{α∈Σ}

consists of a set E of expressions (preliminarily to be thought of as unstructured surface strings), a set A ⊆ E of atomic expressions, and for each function symbol α ∈ Σ a corresponding syntactic rule, which is a partial map \underline{α} from E^{n} to E, for some n.^{1}

• Let V ar be a set of variables, not belonging to E. The set T (E) of terms is defined as follows.

– V ar ∪ E ⊆ T (E)

– If t_{1}, …, t_{n} ∈ T (E) and α ∈ Σ is n-ary, then α(t_{1}, …, t_{n}) ∈ T (E).

T (E) is a set of auxiliary terms which includes (meta)variables and is only introduced to handle substitution conveniently. The interesting terms come next:

• The set GT (E) of grammatical terms and the function val : GT (E) ⟶ E are given by:

– a ∈ A is an atomic grammatical term, and val(a) = a.

– Suppose α ∈ Σ is n-ary, and that p_{1}, …, p_{n} ∈ GT (E) with val(p_{i}) = e_{i}. If \underline{α}(e_{1}, …, e_{n}) is defined, the term α(p_{1}, …, p_{n}) is in GT (E), and val(α(p_{1}, …, p_{n})) = \underline{α}(e_{1}, …, e_{n}).

If val(p) = e then p is a structural analysis of e.

• A semantics for E is a partial function μ from GT (E) to some set M of meanings. p ∈ GT (E) is μ-meaningful if p ∈ dom(μ), and p and q are μ-synonymous, p ≡_{μ} q, if μ(p) = μ(q). ≡_{μ} is an equivalence relation on dom(μ).

If a term is grammatical, then by definition so are its subterms. The corresponding condition for μ-meaningfulness is called the domain principle in [2]. Thus, the domain principle states that if q ∈ dom(μ) and p is a subterm of q, then p ∈ dom(μ). We can now give one formulation of compositionality (there are several more or less equivalent ones in [1]):

5.2 Definition. μ is compositional if it satisfies the domain principle and for each α ∈ Σ there is function r_{α} such that, whenever α(p_{1}, …, p_{n}) is μmeaningful,

μ(α(p_{1}, …, p_{n})) = r_{α}(μ(p_{1}), …, μ(p_{1})).

Syntactic and semantic categories are not given at the outset here, but can be defined as follows: two terms have the same syntactic (semantic) category if they can be substituted everywhere with preserved grammaticality (meaningfulness). More precisely, if s and p are terms, let s(p|x) be the result of replacing all occurrences of the variable x in s by p.

1I use ‘E’ in what follows ambiguously for the grammar and for its set of expressions.

5.3 Definition.

• For all p, q ∈ GT (E),

p ∼_{E} q ⟺ ∀s ∈ T (E)(s(p|x) ∈ GT (E) ⇔ s(q|x) ∈ GT (E)) p ∼_{μ} q ⟺ ∀s ∈ T (E)(s(p|x) ∈ dom(μ) ⇔ s(q|x) ∈ dom(μ)).

∼_{E} and ∼_{μ} are equivalence relations on GT (E). The syntactic (semantic) category of p ∈ GT (E) is its equivalence class [p]_{E} ([p]_{μ}). Cat_{E} is the set of syntactic categories of E.

• E is categorial if, for each α ∈ Σ,

α(p_{1}, …, p_{n}), α(p′_{1}, …, p′_{n}) ∈ GT (E) implies p_{i} ∼_{E} p′_{i}, 1 ≤ i ≤ n.

5.4 Definition. μ is (a) husserlian, (b) weakly husserlian if, correspondingly,

(a) For all p, q ∈ GT (E), p ≡_{μ} q implies p ∼_{μ} q.

(b) For all p, q ∈ GT (E), p ≡_{μ} q implies p ∼_{E} q.

Now we can consider the idiom extension problem. Fix a grammar E, a semantics μ for E, an expression q_{0} = α_{0}(q_{01}, … q_{0k}) ∈ dom(μ) with surface form

e_{0} = val(q_{0}).

Suppose that q_{0} acquires the status of an idiom, which is to mean m_{0}. How can E and μ be extended?

6 Atomic extensions

We noted that one idea (also hinted at in Hodges [2]) is that some idioms are new atoms, looking like familiar expressions on the surface but in fact without syntactic structure. This is easily modeled in the present framework: assume e_{0} ∈ E − A, and consider

E^{a} = (E, A ∪ \{e_{0}\}, \underline{α})_{α∈Σ}.

Thus, the expressions are the same, and so are the grammatical rules. We call E^{a} an atomic extension of E.^{2} Note that T (E^{a}), GT (E^{a}), and val^{a} are now uniquely determined.

6.1 Lemma.

(a) GT (E) = {p ∈ GT (E^{a}) : e_{0} not in p} and val = val^{a}|GT (E).

2Treating kick the bucket as an atom does not necessarily force us to treat (very implausibly) kicked the bucket or kicks the bucket as atoms too. There might be a rule β taking, for instance, John and lift the bucket to a term with value John lifted the bucket, where val(lift the bucket) = e_{1} is not an atom, and the same rule can form the term with value John kicked the bucket, even though val(kick the bucket) = kick the bucket is an atom.

(b) e_{0} ∼_{E^{a}} q_{0}

(c) If s is an e_{0}-free term, then s(e_{0}|x) ∈ GT (E^{a}) ⇔ s(q_{0}|x) ∈ GT (E), and val(s(q_{0}|x)) = val^{a}(s(e_{0}|x)).

(d) If p, q ∈ GT (E), then p ∼_{E} q ⇔ p ∼_{E^{a}} q.

7 Paraphrase semantics

We still have to extend μ. Actually, there are two ways to think about this; the first is considered in this section. Assume then that q_{0} has a paraphrase (for example, die for kick the bucket), i.e., an expression p_{0} ∈ dom(μ) such that p_{0} ∼_{E} q_{0} and μ(p_{0}) = m_{0} is the meaning we want to give to e_{0}.^{3}

Now there is an obvious way to extend μ: just replace e_{0} by p_{0} and use the meaning of the resulting term, whenever defined. Let μ(e_{0}/p_{0}) be the resulting semantics.

7.1 Proposition. If μ is (a) (weakly) husserlian, (b) total, (c) compositional, or (d) obeys the domain principle, then the corresponding property holds for μ(e_{0}/p_{0}).

Observe that μ(e_{0}/p_{0}) is not at all the smallest compositional idiomatic extension of μ. Since e_{0} is an atom, it is possible to just add the meaning of e_{0} and nothing else, preserving compositionality. This is a trivial and uninteresting extension of μ, since it does not allow the idiom as a proper constituent in any meaningful term. In such a semantics, even though the meaning of kick the bucket was determined, the meaning of, say, John kicked the bucket would not be. Instead we have chosen

K(e_{0}/p_{0}) = {s(e_{0}|x) : s(p_{0}|x) ∈ dom(μ)}

as domain for μ(e_{0}/p_{0}). But isn’t this allowing too much, since some rules, such as passivization, do not seem to apply to the idiomatic reading as they did to the literal one? We will see that, as suggested in section 3, independent semantic mechanisms can be used to filter out such terms. These mechanisms would restrict dom(μ) and consequently dom(μ(e_{0}/p_{0})) (but not GT (E^{a})).

Call two semantics for a given grammar equivalent if they have the same associated synonymy relations.

7.2 Proposition. If μ is compositional, then μ(e_{0}/p_{0}) is, up to equivalence, the unique compositional extension of μ which has domain K(e_{0}/p_{0}) and gives e_{0} the meaning μ(p_{0}).

8 How to handle new idiomatic meanings

The strategy from the preceding section does not work if there is no expression of the same syntactic category with exactly the desired idiomatic meaning; pull strings might be a case in point. There is still a clear intuition, I think, that one should be able to easily modify the existing compositional machinery to accommodate the idiom. This intuition can be cashed out in the following way.

3One can see that it would be too much to require that p_{0} ∼_{μ} q_{0}.

Assume now that E is categorial (Def. 5.3). Next, we slightly strengthen the assumption of compositionality.

8.1 Definition.

• A type system for E has the form T = ⟨r_{α}, D_{c}⟩_{α∈Σ,c∈Cat_{E}}. D_{[p]_{E}} is the type of p, and we assume that types are pairwise disjoint, and that for α ∈ Σ, if α(p_{1}, …, p_{n}) ∈ GT (E), r_{α} is a total function

r_{α} : D_{[p_{1}]_{E}} × ⋯ × D_{[p_{n}]_{E}} ⟶ D_{[α(p_{1},...,p_{n})]_{E}}.

(since E is categorial, this is independent of the choice of p_{1}, …, p_{n}).

• T is μ-compositional if μ satisfies the domain principle and, whenever α ∈ Σ, p ∈ GT (E), and α(p_{1}, …, p_{n}) is μ-meaningful,

(i) μ([p]_{E}) ⊆ D_{[p]_{E}} (ii) μ(α(p_{1}, …, p_{n})) = r_{α}(μ(p_{1}), …, μ(p_{n})).

Actually, this strengthening of the notion of compositionality is very slight:

8.2 Proposition. Let μ be a semantics for E. The following are equivalent:

(a) μ is weakly husserlian and compositional.

(b) There is a μ-compositional type system for E.

Now recall our atomic extension E^{a}.

8.3 Lemma.

(a) If E is categorial, so is E^{a}.

(b) If s(e_{0}|x) ∈ GT (E^{a}), [s(e_{0}|x)]_{E^{a}} = [s(q_{0}|x)]_{E^{a}}.

(c) The map ′ defined by ([p]_{E})′ = [p]_{E^{a}} is a bijection from Cat_{E} to Cat_{E^{a}}.

(d) With D_{c′} = D_{c}, T = ⟨r_{α}, D_{c′}⟩_{α∈Σ,c∈Cat_{E}} is a type system for E^{a}.

Now even if the desired meaning m_{0} for our idiom is not in the range of μ, as long as it is in D_{[q_{0}]_{E}} there is an obvious way to use the compositional type system T for E to extend μ inductively to a semantics μ^{a} such that μ^{a}(e_{0}) = m_{0}; we just follow the given meaning operations r_{α}, which works since they are total. The domain of μ^{a} will be

K(e_{0}/q_{0}) = {s(e_{0}|x) : s(q_{0}|x) ∈ dom(μ)}.

Of course this strategy for extending the semantics applies also when we do have a paraphrase; in that case the extensions are the same, as stated in the next result which summarizes our findings so far.

8.4 Theorem. Suppose that E is categorial and that E^{a} is obtained by adding e_{0} ∈ E − A as a new atom, where e_{0} = val(q_{0}). Suppose further that μ is a semantics for E in which q_{0} is meaningful, and that T = ⟨r_{α}, D_{c}⟩_{α∈Σ,c∈Cat_{E}} is a type system for E which is μ-compositional. Finally, suppose m_{0} ∈ D_{[q_{0}]_{E}}. Then the following holds:

(a) There is a unique extension μ^{a} of μ to K(e_{0}/q_{0}) such that T is μ^{a}- compositional and μ^{a}(e_{0}) = m_{0}.

(b) Suppose m_{0} = μ(p_{0}) for some paraphrase p_{0} ∈ dom(μ) such that p_{0} ∼_{E} q_{0}. Then for all p ∈ K(e_{0}/p_{0}) ∩ K(e_{0}/q_{0}), μ(e_{0}/q_{0})(p) = μ^{a}(p).

9 Idioms with syntactic structure

Having some idioms as atoms is just one idea. In other cases, or perhaps always, one may prefer idioms with syntactic structure. Then we still need to distinguish the idiomatic reading to avoid ambiguity, while respecting the intuition that at surface level the idiomatic and the literal versions coincide. There are several ways one could go about this, for our chosen expression q_{0} = α_{0}(q_{01}, … q_{0k}). For example, one could have a special rule/operation which only applied to the arguments (e_{01}, … e_{0k}), where e_{0i} = val(q_{0i}), and was undefined for all others. This treatment, which essentially amounts to just using an idiomatic ‘marker’, is possible although it leads to some problems with categories. Here I will only consider one alternative account, where the idea is to keep using the old operation \underline{α_{0}}, but give it a new name.

This means that our given grammar E will now be extended in the following way.

9.1 Definition.

E^{i} = (E, A, \underline{α})_{α∈Σ^{i}},

where Σ^{i} = Σ∪\{α_{0}^{i}\}, and α_{0}^{i} is a new k-ary function symbol such that \underline{α_{0}^{i}} = \underline{α_{0}}. E^{i} is called a duplicated rule extension of E. Let q_{0}^{i} = α_{0}^{i}(q_{01}, …, q_{0k}).

So now we have a new function symbol, hence new grammatical terms, but the same expressions and the same operations on expressions as before. The relations between E and E^{i} are easily described. For s ∈ T (E^{i}), define

s^{−} = the result of deleting all superscripts ^{i} in s.

Note that s^{−−} = s^{−} and s(p|x)^{−} = s^{−}(p^{−}|x). Also define, for X ⊆ T (E),

X^{+} = {s ∈ T (E^{i}) : s^{−} ∈ X}.

That is, X^{+} consists of all those terms that are obtained from terms in X by adding the superscript ^{i} to zero or more occurrences of α_{0}.

9.2 Lemma.

(a) T (E^{i}) = T (E)^{+} and GT (E^{i}) = GT (E)^{+}.

(b) For p, q ∈ GT (E^{i}), p ∼_{E^{i}} q ⇔ p^{−} ∼_{E} q^{−}.

(c) If p ∈ GT (E^{i}) then [p]_{E^{i}} = [p^{−}]_{E^{i}}, and if p ∈ GT (E) then [p]_{E^{i}} = [p]_{E}^{+}.

(d) If E is categorial, so is E^{i}, and the map ′ from Cat_{E} to Cat_{E^{i}} defined by ([p]_{E})′ = [p]_{E^{i}} is a bijection.

Again, there are two ways to extend the semantics μ. With a paraphrase p_{0} we can do the paraphrase semantics μ(q_{0}^{i}/p_{0}). The domain will be K(q_{0}^{i}/p_{0}) = {s(q_{0}^{i}|x) : s(p_{0}|x) ∈ dom(μ)}, and the same results as before hold.

Alternatively, suppose that E is categorial, and that T = ⟨r_{α}, D_{c}⟩_{α∈Σ,c∈Cat_{E}} is a μ-compositional type system for E. Assume m_{0} ∈ D_{[q_{0}]_{E}}. Letting D_{c′} = D_{c} as before, define a new type system

T^{i} = ⟨r_{α}, D_{c}⟩_{α∈Σ^{i},c∈Cat_{E^{i}}} ,

where r_{α_{0}^{i}} is a function with the same domain as r_{α_{0}}, defined as follows:

r_{α_{0}^{i}}(m_{1}, …, m_{k}) = \{ \begin{matrix}m_{0} & if m_{j} = μ(q_{0j}), 1 ≤ j ≤ k \\ r_{α_{0}}(m_{1}, …, m_{k}) & otherwise. \end{matrix}

Now if α_{0}^{i}(p_{1}, …, p_{k}) ∈ GT (E^{i}) then (α_{0}^{i}(p_{1}, …, p_{k}))^{−} = α_{0}(p_{1}^{−}, …, p_{k}^{−})) ∈ GT (E). Also, D_{[p_{j}^{−}]_{E}} = D_{[p_{j}]_{E^{i}}} and D_{[α_{0}(p_{1}^{−},...,p_{k}^{−})]_{E}} = D_{[α_{0}^{i}(p_{1},...,p_{k})]_{E^{i}}} = D_{[q_{0}]_{E^{i}}}

(since E^{i} is categorial). Hence,

r_{α_{0}^{i}} : D_{[p_{1}]_{E^{i}}} × ⋯ × D_{[p_{n}]_{E^{i}}} ⟶ D_{[α(p_{1},...,p_{n})]_{E^{i}}} .

It follows that T^{i} is a type system for E^{i}. Again, we can use the compositional machinery to extend μ inductively to a semantics μ^{i} for E^{i} such that μ^{i}(q_{0}^{i}) = m_{0}. We will have dom(μ^{i}) = K^{i} = dom(μ)^{+}. The following result holds.

9.3 Theorem. Suppose that E is categorial and that E^{i} is obtained as above by duplicating the last syntactic rule used in forming the term q_{0}. Suppose further that T = ⟨r_{α}, D_{c}⟩_{α∈Σ,c∈Cat_{E}} is a μ-compositional type system for E, and that we wish to give q_{0} the idiomatic meaning m_{0} ∈ D_{[q_{0}]_{E}}. Then the following holds:

(a) There is a unique extension μ^{i} of μ to K^{i} such that T^{i} is μ^{i}-compositional and μ^{i}(q_{0}^{i}) = m_{0}.

(b) Suppose m_{0} = μ(p_{0}) for some paraphrase p_{0} ∈ dom(μ) such that p_{0} ∼_{E} q_{0}. Then for all p ∈ K(q_{0}^{i}/p_{0}) ∩ K^{i}, μ(q_{0}^{i}/p_{0})(p) = μ^{i}(p).

10 Discussion

We have looked at some ways of extending both syntax and semantics in order to incorporate a new idiom. To see that these extensions are at least reasonable, we may observe that criteria like the following have been satisfied:

• Surface identity. On the surface, the idiom looks the same as the nonidiomatic expression.

• Preservation. The extensions preserve various desirable properties, in particular, compositionality. We saw that other properties (such as the husserl property) were also preserved, and we may also note (where E′ and μ′ stand for the extensions that have been considered)

– Preservation of syntactic categories. For p, q ∈ GT (E), p ∼_{E} q implies p ∼_{E′} q.

– Preservation of semantic categories. For p, q ∈ dom(μ), p ∼_{μ} q implies p ∼_{μ′} q.

• Uniqueness. The extension is (in some suitable sense) uniquely determined (given that it should have certain properties).

• Conservativity. If p ∈ GT (E) − dom(μ) then p ∈ GT (E′) − dom(μ′).

• Semantic domain. The extended semantics should have a reasonable and non-trivial domain, allowing the idiom to be used in all contexts one expects to be able to use it.

In the connection with the last item there is also the issue of overgeneration. Let us look at two cases. First, operations such as passivization or anaphoric reference to the object position may not apply to some idioms, like kick the bucket. What happens, on the accounts considered here, with sentences like The bucket was kicked by John? Syntactically, the ‘idiomatic version’ of this sentence will belong to the extended grammar. But semantically we may assume already in the given semantics that, say, passivization does not apply to one-place predicates. That is, applying the passive operation to something like John died will not be meaningful. Then, if we extend by the paraphrase semantics, neither will the ‘idiomatic version’ of The bucket was kicked by John. Thus, regardless of whether we treat the idiom as an atom or as having syntactic structure, passivization can be blocked in the paraphrase semantics by semantic restrictions in the given language. On the other hand, when there is no paraphrase and we just have the syntax and semantics of the literal reading to work with, the idiomatic reading of The bucket was kicked by John as John died will not be blocked (with μ^{a} and μ^{i}). It remains to be seen if there are other means in the present framework to achieve that effect.

Another example concerns the case of John kicked the pail, assuming that the pail and the bucket are synonymous. With the atomic approach, there is no idiomatic reading of this sentence, so the problem does not appear. But suppose we think of kick the bucket as having syntactic structure. Now (an idiomatic version of) John kicked the pail is not in the domain of the paraphrase semantics μ(q_{0}^{i}/p_{0}) (since it is not obtained by replacing die by kick the bucket). But it is meaningful in μ^{i}, and there it will mean John died. Less seriously, a sentence like John lifted the pail will also have an ‘idiomatic version’ (obtained by applying the duplicated function symbol instead), but its meaning will be the ordinary one.

11 Further directions

I have only approached a small number of the problems concerning idioms and compositionality here. An example of something that must be accounted for is the fact that, say, pull strings occurs freely in contexts like

(9) pull all/some/a few/several/many/most/no strings.

But it is still one idiom, not several, so an atomic analysis seems far-fetched, to say the least. The syntactic form is something like

α_{0}(pull, β_{0}(q, strings)),

where q is a determiner. Here the proposed analysis does not apply, and it remains to be seen if it can be suitably adapted.

Also, I have not addressed the proposal in Nunberg et al. [4] that for pull strings we should have the ordinary syntactic rule but idiomatic versions of pull and string instead. It is easy to see the intuitive appeal of this approach, where the meaning of pull strings is computed in the ordinary way from the meanings of new atoms pull^{i} (say, ‘exploit’) and string^{i} (say, ‘connection’). But the notion of atomic extension from the present paper will not work in this case. For the case of kick the bucket, the corresponding surface string is in E − A and so can be taken as a new atom without requiring additional rules, but this does not hold for pull^{i} and string^{i}. They are not in E, so rules have to be extended, but furthermore we took the value of an atom to be itself, and then we cannot account for the ambiguity between pull and pull^{i}. In other words, the framework needs to be adjusted in order to accommodate this proposal. I think such adjustment can be done, but I have not tried to do it here. Note that any treatment of this proposal will have to avoid, in a non-ad hoc way, a more serious kind of overgeneration than what we encountered so far, namely, expressions like pull^{i} the wagon and tie strings^{i}, which appear to make no sense at all.

However, even with these limitations, I hope to have shown that there are interesting problems concerning idioms and compositionality which, when the standard — and sometimes misleading — formulations of these matters have been demystified, can be profitably treated in a general algebraic framework such as the one presented here.

Acknowledgement. This work was partially supported by the Bank of Sweden Tercentenary Foundation in connection with the project Meaning and Interpretation.

References

[1] W. Hodges, Formal features of compositionality, 1998, forthcoming.

[2] W. Hodges, Compositionality is not the problem, Logic and Logical Philosophy 6 (1998), 7-33.

[3] T. Janssen, Compositionality, in J. van Benthem and A. ter Meulen (eds), Handbook of Logic and Language, Elsevier, Amsterdam, 1996, 417–473.

[4] G. Nunberg, I. Sag and T. Wasow, Idioms, Language 70 (1994), 491–538.